C++ Program For Pointing Arbit Pointer To Greatest Value Right Side Node In A Linked List
Given singly linked list with every node having an additional “arbitrary” pointer that currently points to NULL. We need to make the “arbitrary” pointer to the greatest value node in a linked list on its right side.
A Simple Solution is to traverse all nodes one by one. For every node, find the node which has the greatest value on the right side and change the next pointer. The Time Complexity of this solution is O(n2).
An Efficient Solution can work in O(n) time. Below are the steps.
- Reverse the given linked list.
- Start traversing the linked list and store the maximum value node encountered so far. Make arbit of every node to point to max. If the data in the current node is more than the max node so far, update max.
- Reverse modified linked list and return head.
Following is the implementation of the above steps.
C++
// C++ program to point arbit pointers // to highest value on its right #include<bits/stdc++.h> using namespace std; // Link list node struct Node { int data; Node* next, *arbit; }; /* Function to reverse the linked list */ Node* reverse(Node *head) { Node *prev = NULL, *current = head, *next; while (current != NULL) { next = current->next; current->next = prev; prev = current; current = next; } return prev; } // This function populates arbit pointer // in every node to the greatest value // to its right. Node* populateArbit(Node *head) { // Reverse given linked list head = reverse(head); // Initialize pointer to maximum // value node Node *max = head; // Traverse the reversed list Node *temp = head->next; while (temp != NULL) { // Connect max through arbit // pointer temp->arbit = max; // Update max if required if (max->data < temp->data) max = temp; // Move ahead in reversed list temp = temp->next; } // Reverse modified linked list // and return head. return reverse(head); } // Utility function to print result // linked list void printNextArbitPointers(Node *node) { printf ( "Node Next Pointer Arbit Pointer" ); while (node!=NULL) { cout << node->data << " " ; if (node->next) cout << node->next->data << " " ; else cout << "NULL" << " " ; if (node->arbit) cout << node->arbit->data; else cout << "NULL" ; cout << endl; node = node->next; } } /* Function to create a new node with given data */ Node *newNode( int data) { Node *new_node = new Node; new_node->data = data; new_node->next = NULL; return new_node; } // Driver code int main() { Node *head = newNode(5); head->next = newNode(10); head->next->next = newNode(2); head->next->next->next = newNode(3); head = populateArbit(head); printf ( "Resultant Linked List is: " ); printNextArbitPointers(head); return 0; } |
Output:
Resultant Linked List is: Node Next Pointer Arbit Pointer 5 10 10 10 2 3 2 3 3 3 NULL NULL
Time complexity: O(n)
Space complexity: O(1)
Recursive Solution:
We can recursively reach the last node and traverse the linked list from the end. The recursive solution doesn’t require reversing of the linked list. We can also use a stack in place of recursion to temporarily hold nodes. Thanks to Santosh Kumar Mishra for providing this solution.
C++
// C++ program to point arbit pointers // to highest value on its right #include <bits/stdc++.h> using namespace std; // Link list node struct Node { int data; Node *next, *arbit; }; // This function populates arbit pointer // in every node to the greatest value // to its right. void populateArbit(Node* head) { // using static maxNode to keep track // of maximum orbit node address on // right side static Node* maxNode; // if head is null simply return // the list if (head == NULL) return ; /* if head->next is null it means we reached at the last node just update the max and maxNode */ if (head->next == NULL) { maxNode = head; return ; } /* Calling the populateArbit to the next node */ populateArbit(head->next); /* Updating the arbit node of the current node with the maximum value on the right side */ head->arbit = maxNode; /* If current Node value id greater then the previous right node then update it */ if (head->data > maxNode->data) maxNode = head; return ; } // Utility function to print result // linked list void printNextArbitPointers(Node* node) { printf ( "Node Next Pointer Arbit Pointer" ); while (node != NULL) { cout << node->data << " " ; if (node->next) cout << node->next->data << " " ; else cout << "NULL" << " " ; if (node->arbit) cout << node->arbit->data; else cout << "NULL" ; cout << endl; node = node->next; } } /* Function to create a new node with given data */ Node* newNode( int data) { Node* new_node = new Node; new_node->data = data; new_node->next = NULL; return new_node; } // Driver code int main() { Node* head = newNode(5); head->next = newNode(10); head->next->next = newNode(2); head->next->next->next = newNode(3); populateArbit(head); printf ( "Resultant Linked List is: " ); printNextArbitPointers(head); return 0; } |
Output:
Resultant Linked List is: Node Next Pointer Arbit Pointer 5 10 10 10 2 3 2 3 3 3 NULL NULL
Time complexity: O(n) where n is no of nodes in a linked list.
Auxiliary Space: O(1) since using constant space for variables
Please refer complete article on Point arbit pointer to greatest value right side node in a linked list for more details!
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